Answer to Question #116401 in Mechanics | Relativity for Foibe Kambala

Question #116401
7. A 65 kg trampoline artist jumps vertically upward from the top of platform with speed of
5.0 m/s .
a) How fast is he going as he lands on the trampoline 3.0 m below?
b) If the trampoline behaves like a spring with spring stiffness constant 6.2 x 10 4 N/m.
How far does he depress it?
1
Expert's answer
2020-05-21T13:53:20-0400

a)

applying v2=u2+2gSv^2=u^2+2gS

v2=25+2×10×3=95v=9.75m/sv^2=25+2\times10\times3=95\\v=9.75m/s


b)

if the trampoline behaves like a spring then the depression in the spring in the spring will create potential energy

so applying energy conservation

12mv2=12kx2\frac12mv^2=\frac12kx^2

65×(9.75)2=6.2×104×x265\times(9.75)^2=6.2\times10^4\times x^2

x=31.56×102mx=31.56\times10^{-2}m

x=31.56cm


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